Alternating Current, Voltage (rms and Average) Questions in English

(B) For an alternating current $(ac)$ source,the relationship between the peak voltage $(V_0)$ and the root mean square $(rms)$ voltage $(V_{rms})$ is given by the formula: $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Rearranging this formula to solve for the peak voltage,we get: $V_0 = \sqrt{2} \times V_{rms}$.
Therefore,the peak voltage is $\sqrt{2}$ times the $rms$ value of the $ac$ source.

(A) The instantaneous current in an $AC$ circuit is given by $I = I_0 \sin(\omega t)$.
At maximum value,the phase $\omega t_1 = \frac{\pi}{2}$.
At $rms$ value,the current is $I = \frac{I_0}{\sqrt{2}}$,so $\sin(\omega t_2) = \frac{1}{\sqrt{2}}$. The first instance after the peak is $\omega t_2 = \frac{3\pi}{4}$.
The phase difference is $\Delta \phi = \omega t_2 - \omega t_1 = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$.
Given frequency $f = 50 \,Hz$,the angular frequency $\omega = 2\pi f = 2\pi \times 50 = 100\pi \,rad/s$.
The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\pi / 4}{100\pi} = \frac{1}{400} \,s$.
Converting to milliseconds: $t = 0.0025 \,s = 2.5 \,ms$.

(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 5 \text{ A}$ and $\omega = 120 \pi \text{ rad/s}$.
The angular frequency $\omega$ is related to the time period $T$ by the formula $\omega = \frac{2 \pi}{T}$.
Substituting the value of $\omega$: $120 \pi = \frac{2 \pi}{T} \Rightarrow T = \frac{2 \pi}{120 \pi} = \frac{1}{60} \text{ s}$.
The current reaches its peak value at $t = \frac{T}{4}$ (one-quarter of the time period).
Therefore,the time taken is $t = \frac{1/60}{4} = \frac{1}{240} \text{ s}$.

(A) For a domestic $AC$ supply,the given voltage $220 \,V$ represents the root-mean-square $(V_{\text{rms}})$ value.
The peak voltage $(V_{\text{max}})$ is calculated as $V_{\text{max}} = V_{\text{rms}} \times \sqrt{2} = 220 \sqrt{2} \,V$.
The frequency $f$ is $50 \,cps$ (or $50 \,Hz$). The angular frequency $\omega$ is given by $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \,rad/s$.
The instantaneous potential difference $V(t)$ is expressed as $V(t) = V_{\text{max}} \cos(\omega t)$.
Substituting the values,we get $V(t) = 220 \sqrt{2} \cos(100 \pi t) \,V$.

(A) The correct answer is $A$.
$A$ hot wire ammeter is a device used to measure the intensity of $AC$ or $DC$ current.
It operates based on the thermal expansion of a wire that is heated due to the flow of electric current through it.
The heat produced in the wire is proportional to the square of the current $(H \propto I^2 R t)$.
Since the heating effect depends on the square of the current,it is independent of the direction of the current flow.
Therefore,it can measure both the $DC$ and the $rms$ value of $AC$.

(A) In alternating current $(a.c.)$ circuits,the current and voltage vary sinusoidally with time.
Since the average value of an alternating current over a complete cycle is zero,standard $a.c.$ meters (like ammeters and voltmeters) are designed to measure the root mean square $(r.m.s.)$ value.
The $r.m.s.$ value represents the effective value of the alternating current,which is equivalent to the direct current $(d.c.)$ that would produce the same amount of heat in a resistor.
Therefore,the correct option is $A$.

(A) The mean value of current $i_{\text{mean}}$ over a time interval $t_1$ to $t_2$ is given by $i_{\text{mean}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} i(t) dt$.
For the first half cycle ($0$ to $T/2$),the current $i(t)$ represents a straight line passing through the origin with slope $m = \frac{i_0}{T/2} = \frac{2i_0}{T}$.
Thus,$i(t) = \frac{2i_0}{T} t$.
Now,calculating the mean current for the interval $0$ to $T/2$:
$i_{\text{mean}} = \frac{1}{T/2 - 0} \int_{0}^{T/2} \frac{2i_0}{T} t dt$
$i_{\text{mean}} = \frac{2}{T} \cdot \frac{2i_0}{T} \int_{0}^{T/2} t dt$
$i_{\text{mean}} = \frac{4i_0}{T^2} \left[ \frac{t^2}{2} \right]_{0}^{T/2}$
$i_{\text{mean}} = \frac{4i_0}{T^2} \cdot \frac{1}{2} \left( \frac{T^2}{4} - 0 \right)$
$i_{\text{mean}} = \frac{2i_0}{T^2} \cdot \frac{T^2}{4} = \frac{i_0}{2}$.

(A) The heat produced by a heater in a resistance $R$ over time $t$ is given by $H = \frac{V^2}{R} t$.
For $d.c.$ voltage,$H_{dc} = \frac{V_{dc}^2}{R} t$.
For $a.c.$ voltage,the heat produced is determined by the $r.m.s.$ value of the voltage,$H_{ac} = \frac{V_{rms}^2}{R} t$.
Given that the heat produced is the same in both cases $(H_{ac} = H_{dc})$,we have:
$\frac{V_{rms}^2}{R} t = \frac{V_{dc}^2}{R} t$
$V_{rms}^2 = V_{dc}^2$
$V_{rms} = V_{dc}$
Given $V_{dc} = 110 \, V$,therefore $V_{rms} = 110 \, V$.

(D) The rate of heat dissipation (power) in a resistor is given by $P = \frac{V_{rms}^2}{R}$.
For the sinusoidal potential difference $V_1$ with peak value $V_0$,the root-mean-square $(RMS)$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Thus,the rate of heat dissipation is $W = \frac{(V_0 / \sqrt{2})^2}{R} = \frac{V_0^2}{2R}$.
For the square wave potential difference $V_2$ with peak value $V_0$,the voltage alternates between $+V_0$ and $-V_0$. The $RMS$ value for this square wave is $V_{rms} = V_0$.
Thus,the new rate of heat dissipation is $W' = \frac{V_0^2}{R}$.
Comparing the two expressions:
$W' = 2 \times \left( \frac{V_0^2}{2R} \right) = 2W$.
Therefore,the rate of heat dissipation becomes $2W$.

(A) The time period $T$ of the $AC$ source is given by $T = \frac{1}{f} = \frac{1}{60 \,Hz} \approx 0.01667 \,s = 16.67 \,ms$.
The average value of an alternating voltage $V(t) = V_m \sin(\omega t)$ over one complete cycle (time interval $T$) is given by $V_{avg} = \frac{1}{T} \int_{0}^{T} V_m \sin(\omega t) \,dt$.
Since the integral of a sine function over one complete period is zero,the average voltage over $16.67 \,ms$ must be zero.

(B) The total current $I$ flowing through the wire is given by the sum of the direct current and the alternating current: $I = 5 + 10 \sin \omega t$.
The effective value ($RMS$ value) of the current is defined as $I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$.
Substituting the expression for $I$: $I_{\text{eff}} = \sqrt{\frac{1}{T} \int_0^T (5 + 10 \sin \omega t)^2 dt}$.
Expanding the square: $(5 + 10 \sin \omega t)^2 = 25 + 100 \sin^2 \omega t + 100 \sin \omega t$.
Integrating over one complete cycle $(T)$:
$1$. The integral of $25$ over $T$ is $25T$.
$2$. The integral of $100 \sin^2 \omega t$ over $T$ is $100 \times (T/2) = 50T$.
$3$. The integral of $100 \sin \omega t$ over $T$ is $0$.
Thus,$I_{\text{eff}} = \sqrt{\frac{1}{T} (25T + 50T + 0)} = \sqrt{25 + 50} = \sqrt{75} = 5\sqrt{3}\,A$.

(A) The power consumed by the lamp is given by $P = V_{rms} \times I_{rms}$.
Given $P = 100\,W$ and $V_{rms} = 200\,V$.
Therefore,the root mean square current $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{200} = 0.5\,A$.
The peak current $I_{peak}$ is related to the $RMS$ current by the formula $I_{peak} = I_{rms} \times \sqrt{2}$.
Substituting the value of $I_{rms}$,we get $I_{peak} = 0.5 \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.707\,A$.

(D) The given current is $i = I_{dc} + I_{ac} \sin(\omega t + \phi)$,where $I_{dc} = 6 \ A$ and the peak value of the alternating component is $I_m = \sqrt{56} \ A$.
The root mean square (rms) value of a composite current is given by the formula $I_{\text{rms}} = \sqrt{I_{dc}^2 + \frac{I_m^2}{2}}$.
Substituting the values:
$I_{\text{rms}} = \sqrt{6^2 + \frac{(\sqrt{56})^2}{2}}$
$I_{\text{rms}} = \sqrt{36 + \frac{56}{2}}$
$I_{\text{rms}} = \sqrt{36 + 28}$
$I_{\text{rms}} = \sqrt{64}$
$I_{\text{rms}} = 8 \ A$.

(C) The potential difference between terminals $X$ and $Y$ is $V_{XY} = V_x - V_y = V_0 [\sin \omega t - \sin(\omega t + 2\pi/3)]$.
Using the formula $\sin A - \sin B = 2 \sin((A-B)/2) \cos((A+B)/2)$,we get:
$V_{XY} = V_0 [2 \sin(-\pi/3) \cos(\omega t + \pi/3)] = V_0 [2 \cdot (-\sqrt{3}/2) \cos(\omega t + \pi/3)] = -\sqrt{3} V_0 \cos(\omega t + \pi/3) = \sqrt{3} V_0 \sin(\omega t + \pi/3 - \pi/2) = \sqrt{3} V_0 \sin(\omega t - \pi/6)$.
The peak value of $V_{XY}$ is $\sqrt{3} V_0$. The $rms$ value is $V_{XY}^{rms} = \frac{\sqrt{3} V_0}{\sqrt{2}} = V_0 \sqrt{\frac{3}{2}}$.
Similarly,for $V_{YZ} = V_y - V_z$,the peak value is also $\sqrt{3} V_0$,so $V_{YZ}^{rms} = V_0 \sqrt{\frac{3}{2}}$.
Thus,both $V_{XY}^{rms}$ and $V_{YZ}^{rms}$ are equal to $V_0 \sqrt{\frac{3}{2}}$. Therefore,options $A$ and $D$ are correct.

(C) The given current is $I = I_A \sin \omega t + I_B \cos \omega t$.
We can rewrite this as $I = \sqrt{I_A^2 + I_B^2} \sin(\omega t + \phi)$,where $\tan \phi = \frac{I_B}{I_A}$.
This represents a sinusoidal current with peak value $I_0 = \sqrt{I_A^2 + I_B^2}$.
The r.m.s. value of a sinusoidal current $I = I_0 \sin(\omega t + \phi)$ is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$,we get $I_{rms} = \frac{\sqrt{I_A^2 + I_B^2}}{\sqrt{2}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}$.

(A) The power rating of the bulb is $P = 100 \ W$ and the voltage rating is $V_{rms} = 220 \ V$.
Using the formula for power in an $ac$ circuit,$P = V_{rms} \times I_{rms}$.
Therefore,the $rms$ current is $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{220} \approx 0.4545 \ A$.
The peak current $I_0$ is related to the $rms$ current by the relation $I_0 = \sqrt{2} \times I_{rms}$.
Substituting the values,$I_0 = 1.414 \times 0.4545 \approx 0.64 \ A$.

(C) The standard equation for alternating current is $i = i_0 \sin(\omega t)$.
Comparing this with the given equation $i = 100 \sqrt{2} \sin(100 \pi t)$,we get the peak current $i_0 = 100 \sqrt{2} \ A$ and angular frequency $\omega = 100 \pi \ rad/s$.
The $\text{RMS}$ value of current is given by $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 \ A$.
The frequency $f$ is given by $f = \frac{\omega}{2 \pi} = \frac{100 \pi}{2 \pi} = 50 \ Hz$.
Thus,the $\text{RMS}$ value is $100 \ A$ and the frequency is $50 \ Hz$.

(C) The given current is $i = 5 \sqrt{2} + 10 \cos \left(650 \pi t + \frac{\pi}{6}\right)$.
To find the $r.m.s$ value,we use the formula $I_{rms} = \sqrt{\langle i^2 \rangle}$.
First,calculate $i^2 = \left(5 \sqrt{2} + 10 \cos \left(650 \pi t + \frac{\pi}{6}\right)\right)^2$.
$i^2 = (5 \sqrt{2})^2 + (10 \cos \left(650 \pi t + \frac{\pi}{6}\right))^2 + 2(5 \sqrt{2})(10 \cos \left(650 \pi t + \frac{\pi}{6}\right))$.
$i^2 = 50 + 100 \cos^2 \left(650 \pi t + \frac{\pi}{6}\right) + 100 \sqrt{2} \cos \left(650 \pi t + \frac{\pi}{6}\right)$.
Taking the average over a full cycle,$\langle \cos \theta \rangle = 0$ and $\langle \cos^2 \theta \rangle = \frac{1}{2}$.
$\langle i^2 \rangle = 50 + 100 \left(\frac{1}{2}\right) + 0 = 50 + 50 = 100$.
Therefore,$I_{rms} = \sqrt{100} = 10 \text{ A}$.

(B) The given current is $i = 3 + 4 \sin(\omega t + \pi/3)$.
To find the $r.m.s.$ value,we use the formula $I_{rms} = \sqrt{\langle i^2 \rangle}$.
First,calculate $i^2 = (3 + 4 \sin(\omega t + \pi/3))^2 = 9 + 16 \sin^2(\omega t + \pi/3) + 24 \sin(\omega t + \pi/3)$.
Taking the average over a full cycle,$\langle 9 \rangle = 9$,$\langle \sin^2(\omega t + \pi/3) \rangle = 1/2$,and $\langle \sin(\omega t + \pi/3) \rangle = 0$.
Thus,$\langle i^2 \rangle = 9 + 16(1/2) + 24(0) = 9 + 8 = 17$.
Therefore,$I_{rms} = \sqrt{17} \ A$.

(B) $(1)$ For $x = x_0 \sin \omega t$, the peak value is $x_0$. The $\text{r.m.s.}$ value is $\frac{x_0}{\sqrt{2}}$. Thus, $(A \rightarrow ii)$.
$(2)$ For $x = x_0 \sin \omega t \cos \omega t = \frac{x_0}{2} \sin(2 \omega t)$, the peak value is $\frac{x_0}{2}$. The $\text{r.m.s.}$ value is $\frac{x_0/2}{\sqrt{2}} = \frac{x_0}{2 \sqrt{2}}$. Thus, $(B \rightarrow iii)$.
$(3)$ For $x = x_0 \sin \omega t + x_0 \cos \omega t = x_0 \sqrt{2} \sin(\omega t + \pi/4)$, the peak value is $x_0 \sqrt{2}$. The $\text{r.m.s.}$ value is $\frac{x_0 \sqrt{2}}{\sqrt{2}} = x_0$. Thus, $(C \rightarrow i)$.
Therefore, the correct matching is $(A \rightarrow ii), (B \rightarrow iii), (C \rightarrow i)$.

(C) The instantaneous current is given by $I = 3 \sin \left(50 \pi t + \frac{\pi}{4}\right)$.
For the current to be maximum,the sine function must be equal to $1$,which means the argument of the sine function must be $\frac{\pi}{2}$.
So,$50 \pi t + \frac{\pi}{4} = \frac{\pi}{2}$.
Subtracting $\frac{\pi}{4}$ from both sides,we get $50 \pi t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.
Dividing by $50 \pi$,we get $t = \frac{\pi}{4 \times 50 \pi} = \frac{1}{200} \text{ s}$.

(C) The given equation for alternating e.m.f. is $e = e_0 \sin \omega t$.
We want to find the time $t$ when $e = e_0/2$.
Substituting this into the equation: $e_0/2 = e_0 \sin \omega t$.
This simplifies to $\sin \omega t = 1/2$.
Since $\sin 30^{\circ} = 1/2$ and $30^{\circ} = \pi/6$ radians,we have $\omega t = \pi/6$.
We know that $\omega = 2\pi/T$,where $T$ is the time period.
Substituting $\omega$ into the equation: $(2\pi/T) \cdot t = \pi/6$.
Solving for $t$: $t = (\pi/6) \cdot (T/2\pi) = T/12$.
Thus,the time taken is $T/12$.

(D) The given equation for the alternating current is $I = 100 \sin(200 \pi t)$.
Comparing this with the standard form $I = I_0 \sin(\omega t)$,the peak value $I_0$ is $100 \ A$.
The current reaches its peak value when $\sin(200 \pi t) = 1$.
This occurs when the argument of the sine function is equal to $\frac{\pi}{2}$.
So,$200 \pi t = \frac{\pi}{2}$.
Dividing both sides by $\pi$,we get $200 t = \frac{1}{2}$.
Therefore,$t = \frac{1}{2 \times 200} = \frac{1}{400} \ s$.

(A) The given alternating current equation is $I = I_0 \sin(\omega t)$.
Comparing this with the given equation $I = 100 \sin(50 \pi t)$, we get angular frequency $\omega = 50 \pi \text{ rad/s}$.
We know that $\omega = 2 \pi f$, where $f$ is the frequency.
So, $2 \pi f = 50 \pi$, which gives $f = 25 \text{ Hz}$.
This means the current completes $25$ cycles in one second.
In one complete cycle of a sine wave, the current becomes zero twice (at the start/end and at the half-cycle point).
Therefore, in $25$ cycles, the current will become zero $25 \times 2 = 50$ times in one second.

(D) For the voltage to be maximum,the sine function must equal $1$.
$\sin \left(\omega t+\frac{\pi}{3}\right) = 1$
Since $\sin \left(\frac{\pi}{2}\right) = 1$,we have:
$\omega t + \frac{\pi}{3} = \frac{\pi}{2}$
$\omega t = \frac{\pi}{2} - \frac{\pi}{3}$
$\omega t = \frac{\pi}{6}$
Substituting $\omega = \frac{2\pi}{T}$:
$\left(\frac{2\pi}{T}\right) t = \frac{\pi}{6}$
$t = \frac{\pi}{6} \times \frac{T}{2\pi}$
$t = \frac{T}{12}$

(B) The instantaneous e.m.f. is given by $e = 200 \sin(50t) \text{ V}$.
Comparing this with the standard equation $e = e_0 \sin(\omega t)$, we get the peak e.m.f. $e_0 = 200 \text{ V}$.
The peak current $I_0$ is calculated using Ohm's law: $I_0 = \frac{e_0}{R} = \frac{200}{50} = 4 \text{ A}$.
The r.m.s. value of current $I_{\text{rms}}$ is given by the formula $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$.
Substituting the values: $I_{\text{rms}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} \text{ A}$.
Since $\sqrt{2} \approx 1.414$, we have $I_{\text{rms}} = 2 \times 1.414 = 2.828 \text{ A}$.

(A) The phase of the current is $\phi_I = \omega t - \pi / 6$.
The phase of the voltage is $\phi_E = \omega t + \pi / 3$.
The phase difference between voltage and current is given by $\Delta \phi = \phi_E - \phi_I$.
Substituting the values,we get $\Delta \phi = (\omega t + \pi / 3) - (\omega t - \pi / 6)$.
$\Delta \phi = \pi / 3 + \pi / 6 = 2\pi / 6 + \pi / 6 = 3\pi / 6 = \pi / 2$.
Since $\Delta \phi$ is positive,the voltage leads the current by $\pi / 2$.

(B) The power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Given $P = 1000 \,W$ and the peak voltage $V_0 = 200 \,V$.
The $r.m.s.$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} \,V$.
Assuming the lamps are purely resistive, the phase angle $\phi = 0^{\circ}$, so $\cos \phi = 1$.
Thus, $1000 = \left( \frac{200}{\sqrt{2}} \right) I_{rms} \times 1$.
$I_{rms} = \frac{1000 \times \sqrt{2}}{200} = 5 \sqrt{2} \,A$.

(D) The instantaneous current is given by $I = 6 \sin(100 \pi t + \frac{\pi}{4})$.
The instantaneous voltage is given by $V = 5 \sin(100 \pi t - \frac{\pi}{4})$.
The phase of the current is $\phi_I = 100 \pi t + \frac{\pi}{4}$.
The phase of the voltage is $\phi_V = 100 \pi t - \frac{\pi}{4}$.
The phase difference between current and voltage is $\Delta \phi = \phi_I - \phi_V = (100 \pi t + \frac{\pi}{4}) - (100 \pi t - \frac{\pi}{4}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.
Since $\Delta \phi = \frac{\pi}{2}$ (or $90^{\circ}$) and the phase of the current is greater than the phase of the voltage,the current leads the voltage by $90^{\circ}$.

(D) The average value of an $A$.$C$. voltage $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is calculated as follows:
$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V dt}{\int_{0}^{\frac{\pi}{\omega}} dt}$
$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V_{m} \sin(\omega t) dt}{\frac{\pi}{\omega} - 0}$
$V_{av} = \frac{V_{m}}{\frac{\pi}{\omega}} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{\frac{\pi}{\omega}}$
$V_{av} = \frac{V_{m} \omega}{\pi} \left( -\frac{1}{\omega} \right) [\cos(\pi) - \cos(0)]$
$V_{av} = -\frac{V_{m}}{\pi} [-1 - 1]$
$V_{av} = -\frac{V_{m}}{\pi} [-2] = \frac{2 V_{m}}{\pi}$

(A) The phase of the current is $\phi_i = -\frac{\pi}{6}$.
The phase of the voltage is $\phi_E = +\frac{\pi}{3}$.
The phase difference $\phi$ by which the voltage leads the current is given by $\phi = \phi_E - \phi_i$.
Substituting the values,we get $\phi = \frac{\pi}{3} - (-\frac{\pi}{6})$.
$\phi = \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi + \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.
Therefore,the voltage leads the current by $\frac{\pi}{2}$.

(A) Let $t^{\prime}$ be the time when the e.m.f. is half its maximum value.
Given the equation $e = e_0 \sin \omega t$,we set $e = \frac{e_0}{2}$.
$\frac{e_0}{2} = e_0 \sin (\omega t^{\prime})$
$\frac{1}{2} = \sin (\omega t^{\prime})$
Since $\sin 30^{\circ} = 0.5$,we have $\omega t^{\prime} = \frac{\pi}{6}$.
Substituting $\omega = \frac{2\pi}{T}$,we get:
$(\frac{2\pi}{T}) t^{\prime} = \frac{\pi}{6}$
$t^{\prime} = \frac{T}{12}$.

(B) The given equation is $V = 80 \sin(100 \pi t) \cos(100 \pi t)$.
Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we can rewrite the equation as:
$V = 40 \times (2 \sin(100 \pi t) \cos(100 \pi t))$
$V = 40 \sin(200 \pi t)$.
Comparing this with the standard form of alternating voltage $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage,
we get $V_0 = 40 \text{ V}$.

(D) The given equation for the instantaneous current is $I = 50 \sin(100 \pi t)$.
We need to find the time $t$ when $I = 25 \ A$.
Substituting the values into the equation:
$25 = 50 \sin(100 \pi t)$
$\frac{25}{50} = \sin(100 \pi t)$
$0.5 = \sin(100 \pi t)$
Since $\sin 30^{\circ} = 0.5$ and $30^{\circ} = \frac{\pi}{6}$ radians,we have:
$100 \pi t = \frac{\pi}{6}$
Dividing both sides by $100 \pi$:
$t = \frac{\pi}{6 \times 100 \pi}$
$t = \frac{1}{600} \ s$.

(D) The given equation for alternating emf is $e = e_0 \cos \omega t$.
Given peak value $e_0 = 10 \ V$ and frequency $f = 50 \ Hz$.
The angular frequency is $\omega = 2 \pi f = 2 \pi \times 50 = 100 \pi \ rad/s$.
Substituting the values into the equation at $t = \frac{1}{600} \ s$:
$e = 10 \cos(100 \pi \times \frac{1}{600})$
$e = 10 \cos(\frac{\pi}{6})$
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we get:
$e = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ V$.

(A) Given the alternating e.m.f. equation: $e = e_0 \sin \omega t$.
We want to find the time $t$ when $e = \frac{e_0}{2}$.
Substituting the value into the equation: $\frac{e_0}{2} = e_0 \sin \omega t$.
This simplifies to: $\sin \omega t = \frac{1}{2}$.
Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} \text{ radians}$.
Substituting $\omega = \frac{2\pi}{T}$: $\left( \frac{2\pi}{T} \right) t = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{T}{12}$.

(A) The given equation for the alternating e.m.f. is $e = e_{0} \sin(\omega t)$.
We know that $\omega = \frac{2\pi}{T}$,so the equation becomes $e = e_{0} \sin\left(\frac{2\pi t}{T}\right)$.
We want to find the time $t$ when $e = \frac{e_{0}}{2}$.
Substituting this into the equation: $\frac{e_{0}}{2} = e_{0} \sin\left(\frac{2\pi t}{T}\right)$.
$\frac{1}{2} = \sin\left(\frac{2\pi t}{T}\right)$.
Since $\sin(30^{\circ}) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\frac{2\pi t}{T} = \frac{\pi}{6}$.
Solving for $t$: $t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12}$.

(C) Given frequency $f = 50 \,Hz$. The time period $T$ is given by $T = \frac{1}{f} = \frac{1}{50} = 0.02 \,s$.
The time taken to reach from zero to the maximum value (peak) is one-quarter of the time period:
$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \,s$.
The peak current $I_0 = 14.14 \,A$. The r.m.s. value of the current $I_{rms}$ is given by:
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{14.14}{1.414} = 10 \,A$.
Thus,the time taken is $0.005 \,s$ and the r.m.s. value is $10 \,A$.

(C) Given,the alternating voltage is $E = 100 \sin \left(\omega t + \frac{\pi}{6}\right) \text{ V}$.
For the voltage to be maximum,the sine term must be equal to $1$.
$\sin \left(\omega t + \frac{\pi}{6}\right) = 1$
Since $\sin \frac{\pi}{2} = 1$,we have:
$\omega t + \frac{\pi}{6} = \frac{\pi}{2}$
$\omega t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$
Using the relation $\omega = \frac{2\pi}{T}$,we substitute $\omega$:
$\left(\frac{2\pi}{T}\right) t = \frac{\pi}{3}$
$t = \frac{\pi}{3} \times \frac{T}{2\pi} = \frac{T}{6}$
Thus,the voltage will be maximum for the first time at $t = \frac{T}{6}$.

(A) The alternating current is represented by the equation $I = I_0 \sin(\omega t)$.
To find the average value over a complete cycle of period $T$,we integrate the current over the interval $[0, T]$ and divide by the total time $T$.
$I_{avg} = \frac{1}{T} \int_{0}^{T} I_0 \sin(\omega t) dt$.
Since $\omega = \frac{2\pi}{T}$,the integral of $\sin(\omega t)$ over a full period is zero because the positive area of the sine wave cancels out the negative area.
Therefore,the average value of alternating current over a complete cycle is $0$.

(B) The instantaneous current is given by the equation $I = I_{0} \sin (\omega t + \phi)$,where $I_{0}$ is the peak current.
Comparing this with the given equation $I = 5 \sin (\omega t + \phi)$,we find the peak current $I_{0} = 5 \text{ A}$.
The root mean square $(rms)$ value of an alternating current is related to the peak current by the formula $I_{rms} = \frac{I_{0}}{\sqrt{2}}$.
Substituting the value of $I_{0}$,we get $I_{rms} = \frac{5}{\sqrt{2}} \text{ A}$.

(D) An alternating quantity varies periodically with time. One complete set of positive and negative values of an alternating quantity is defined as a $cycle$. It represents the complete variation of the quantity before it starts repeating itself.

(B) The root mean square $(I_{\text{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantaneous current over one complete cycle.
For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$,the $I_{\text{rms}}$ value is calculated as:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I^{2} dt} = \frac{I_{0}}{\sqrt{2}}$
where $I_{0}$ is the peak value (amplitude) of the current.

(D) An alternating voltage is defined as a voltage that changes its magnitude and direction periodically with time. The standard mathematical expression for an alternating voltage is $V(t) = V_m \sin(\omega t)$,where $V_m$ is the peak voltage and $\omega$ is the angular frequency. Since this expression follows a sine function,the alternating voltage varies sinusoidally with time.

(B) The standard frequency of alternating current $(AC)$ mains supplied in India is $50 \text{ Hz}$.
Frequency is defined as the number of cycles per second,which is equivalent to $50 \text{ cycles/second}$ or $50 \text{ s}^{-1}$.

(B) $AC$ measuring instruments,such as $AC$ ammeters and voltmeters,are designed to measure the root mean square $(rms)$ value of the alternating current or voltage. This is because the heating effect of current,which is the principle behind most analog measuring instruments,is proportional to the square of the current,making the $rms$ value the most physically significant quantity for power calculations.

(B) The output power of the transformer is given by $P_s = V_s I_s$,where $V_s$ is the $RMS$ voltage and $I_s$ is the $RMS$ current.
Given $P_s = 12 \ W$ and $V_s = 24 \ V$.
Therefore,the $RMS$ current $I_s = \frac{P_s}{V_s} = \frac{12}{24} = 0.5 \ A$.
The peak current $I_m$ is related to the $RMS$ current $I_s$ by the formula $I_m = \sqrt{2} I_s$.
Substituting the value of $I_s$,we get $I_m = \sqrt{2} \times 0.5 \approx 1.414 \times 0.5 = 0.707 \ A$.
Rounding to two decimal places,the peak current is $0.71 \ A$.

(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.
Comparing this with the given equation $I = 50 \cos(100t + 45^{\circ}) \ A$,we find the peak current $I_m = 50 \ A$.
The root mean square current $I_{rms}$ is related to the peak current $I_m$ by the formula $I_{rms} = \frac{I_m}{\sqrt{2}}$.
Substituting the value of $I_m$,we get $I_{rms} = \frac{50}{\sqrt{2}}$.
To simplify,multiply the numerator and denominator by $\sqrt{2}$: $I_{rms} = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2} \ A$.
Therefore,the correct option is $D$.

(B) The resultant current is given by $I = 8 + 6 \sin \omega t$.
To find the $rms$ value,we first calculate the mean square value $\langle I^2 \rangle$.
$\langle I^2 \rangle = \langle (8 + 6 \sin \omega t)^2 \rangle$
$\langle I^2 \rangle = \langle 64 + 96 \sin \omega t + 36 \sin^2 \omega t \rangle$
Using the properties of average values over a full cycle: $\langle \sin \omega t \rangle = 0$ and $\langle \sin^2 \omega t \rangle = \frac{1}{2}$.
$\langle I^2 \rangle = 64 + 96(0) + 36(\frac{1}{2})$
$\langle I^2 \rangle = 64 + 18 = 82 \ A^2$
$I_{rms} = \sqrt{\langle I^2 \rangle} = \sqrt{82} \approx 9.05 \ A$.

(A) The $rms$ value of the alternating current is given by $I_{rms} = \frac{I_m}{\sqrt{2}}$.
Given $I_m = 5 \ A$,we have $I_{rms} = \frac{5}{1.414} \approx 3.536 \ A$.
The time taken to reach the peak value from zero is one-fourth of the time period $T$.
Since $T = \frac{1}{f} = \frac{1}{60} \ s$,the time $t = \frac{T}{4} = \frac{1}{4 \times 60} \ s$.
$t = \frac{1}{240} \ s \approx 0.004167 \ s = 4.167 \ ms$.